The next thing we need is the distance $D$ from the intersection point to CGS. The same page also has a calculator for that and gives $ D \approx 4.168 \, \text{km} \approx 2.251 \, \text{NM} $.

Assuming 140 kt TAS, this gives a turn radius of $ R \approx 4522 \, \mathrm{ft} \approx 0.7442 \, \mathrm{NM} $

The last thing missing is the distance $d$ before the intersection point, where the turn should be started. With a bit of geometry, one finds that this is given by

The next thing we need is the distance $D$ from the intersection point to CGS. The same page also has a calculator for that and gives $ D \approx 4.168 \, \text{km} \approx 2.251 \, \text{NM} $.

Assuming 140 kt TAS, this gives a turn radius of $ R \approx 4522 \, \mathrm{ft} \approx 0.7442 \, \mathrm{NM} $. The corresponding bank angle is then given by (source):

$$ b = 57.3^\circ \arctan \left( \frac{V [\text{kt}]}{362.1 \, \text{kt}} \right) \approx 21.14^\circ $$

The last thing missing is the distance $d$ before the intersection point, where the turn should be started. With a bit of geometry, one finds that this is given by

If you really want to calculate the point to start turning, consider the following sketch:

The first thing we need is the intersection point of the track 354° from MN04A and the extended runway centerline (I chose runway 04L here because that is the default runway for landing in Nice). I looked up the coordinates for the relevant points in my navigation database:

You can use the calculator on movable-type.co.uk to figure out the coordinates for the intersection point: 43° 36′ 28″ N, 007° 08′ 50″ E.

The next thing we need is the distance $D$ from the intersection point to CGS. The same page also has a calculator for that and gives $ D \approx 4.168 \, \text{km} \approx 2.251 \, \text{NM} $.

Now we calculate the turn radius $R$ using the formulas from this answer:

$$ R [\text{ft}] = \frac{V [\text{kt}]^2}{11.26 \tan \theta} \; , \qquad \omega [^\circ/\text{s}] = \frac{1 \, 091 \tan \theta}{V [\text{kt}]} $$

When flying a standard turn at 3 degrees per second, we can solve the second equation for $\tan \theta$ and insert it into the first one:

$$ R [\text{ft}] = \frac{V [\text{kt}]^2}{11.26 \, \omega [^\circ/\text{s}] \, V [\text{kt}] / 1 \, 091} \approx 96.89 \frac{V [\text{kt}]}{\omega [^\circ/\text{s}]} $$

Assuming 140 kt TAS, this gives a turn radius of $ R \approx 4522 \, \mathrm{ft} \approx 0.7442 \, \mathrm{NM} $

The last thing missing is the distance $d$ before the intersection point, where the turn should be started. With a bit of geometry, one finds that this is given by

$$ d = R \tan \left( \frac{\alpha}{2} \right) $$

with our known $ \alpha = 49^\circ $. This results in $ d \approx 0.3392 \, \text{NM} $.

This gives the final answer for the distance from CGS where the turn should be started:

$$ D + d \approx 2.59 \, \text{NM} $$

Note that this entire calculation does not take wind into account!