The figure shows a thick spherical metal shell, with inner radius 4.0 cm and outer radius 6.0 cm…

(a) Inner surface. A closed surface inside the metal has zero flux, since $E = 0$ there. So it encloses no net charge: the inner surface carries $-q = -5.0\ \text{nC}$.

(b) Outer surface. The shell as a whole has $+2.0\ \text{nC}$. So $+2.0\ \text{nC} = (-5.0\ \text{nC}) + Q_{out}$, which gives $Q_{out} = 7.0\ \text{nC}$.

(c) Field at Y. Outside the shell, only the total charge $5.0 + 2.0 = 7.0\ \text{nC}$ matters. $E = \dfrac{kQ}{r^2} = \dfrac{9.0\times10^{9} \times 7.0\times10^{-9}}{(0.10)^2} = 6300\ \text{N C}^{-1}$.

(d) Potential at X. The shell is at $V_s = \dfrac{k \times 7.0\ \text{nC}}{0.060\ \text{m}} = 1050\ \text{V}$. From the shell inwards to X, only the point charge sets the field, so $V_X = V_s + kq\left(\dfrac{1}{r} - \dfrac{1}{a}\right) = 1050 + 45 \times (50 - 25) = 2175\ \text{V}$.

(e) After earthing. The shell is now at $0\ \text{V}$, so its outer surface has no charge (zero charge on it, zero potential). The inner surface still has $-5.0\ \text{nC}$, as before. So $V_X = 0 + 45 \times 25 = 1125\ \text{V}$.