A point charge +q is held at a distance d in front of a very large grounded conducting plane.

Use the method of images: an image $-q$ at distance $2d$ behind the plane reproduces the field in front. (A) is correct: $F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{(2d)^2}=\dfrac{q^2}{16\pi\varepsilon_0d^2}$. (B) is correct: all field lines from $+q$ end on the plane, so the induced charge totals $-q$. (C) is incorrect: the force at separation $x$ from the plane is $q^2/(16\pi\varepsilon_0x^2)$, so $W=\int_d^\infty\dfrac{q^2}{16\pi\varepsilon_0x^2}dx=\dfrac{q^2}{16\pi\varepsilon_0d}$, which is half of the stated value (the value $q^2/(8\pi\varepsilon_0 d)$ would be the energy of a real pair of charges $q$ and $-q$ separated by $2d$, not the work against the induced charge). (D) is correct: at the foot point the two charges give a field $2\cdot\dfrac{q}{4\pi\varepsilon_0d^2}$ normal to the plane, and $\sigma=-\varepsilon_0E=-\dfrac{q}{2\pi d^2}$.