First, use the motor nameplate data to find kva (apparent power). Subscript 3 below indicates 3 phase.
$$ S_3 = \frac{HP * 746}{power factor * efficiency}$$
But, in your case the nameplate gives rated current as 48A, so
$$ S_3 = 48A * 575 * \sqrt3 = 47.8kva$$
$$ P_3 = S_3 * power factor = 47.8 * 0.80 = 38.24kW $$
So,
$$ Q_3 = \sqrt{S_3^2-P_3^2} = 28.7kvar $$
Now, to get to your desired 0.95 power factor you need to provide some of that kvar from your capacitor.
$$ {desired power factor angle} = cos^-{^1}(0.95) = 18.2⁰$$
$$18.2⁰ = tan^-{^1}(Q_3/P_3)$$
$$18.2⁰= tan^-{^1}(Q_{target}/38.24kW) $$
So,
$$Q_{target} = 12.6kvar $$
That means you need a capacitor that will supply an additional (28.7-12.6) = 16.1kvar 3-phase to the motor so that only 12.6kvar comes from the source.
Since,
$$X_C = \frac{V^2}{Q}$$
$$X_C = \frac{575^2}{16.1kvar} = 20.5Ω $$
From this, we can find C (assuming 60Hz) as
$$ C = \frac{1}{2*π*f*X_C} = 129uF $$
EDIT: I noticed that bxjockey had given us nameplate amps so we can directly calculate apparent power. That changed the subsequent results which i have now corrected.