A parallel beam of light travelling in air (refractive index 1.0)...

For refraction at a single spherical surface, we use the formula: $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$
Given data:
Refractive index of air $\mu_1 = 1.0$
Refractive index of glass $\mu_2 = 1.5$
Radius of curvature $R = +50$ cm (convex surface)
Object is at infinity for a parallel beam, so $u = -\infty$
Substituting the values:
$\frac{1.5}{v} - \frac{1.0}{-\infty} = \frac{1.5 - 1.0}{50}$
$\frac{1.5}{v} - 0 = \frac{0.5}{50}$
$\frac{1.5}{v} = \frac{1}{100}$
$v = 1.5 \times 100 = 150$ cm
The image is formed at a distance of $150$ cm from the pole (vertex) of the spherical surface.
The question asks for the distance $x$ from the centre of curvature.
Distance from centre of curvature $x = |v - R|$
$x = 150 - 50 = 100$ cm