Answer
a) Want: $V \propto \Delta(p)$?
$B = -\tfrac{V_o\Delta(p)}{\Delta(V)} \implies -B\tfrac{\Delta(V)}{V_o} = \Delta(p)$
$\Leftrightarrow p_f - p_o = -B(\tfrac{V-V_o}{V_o})\Leftrightarrow p_o - p_f = -B(\tfrac{V-V_o}{V_o})$ $\Leftrightarrow \tfrac{V_o}{B}(p_o - p_f) +V_o = V \Leftrightarrow V = -k V_o\Delta(p) +V_o \because k = \tfrac{1}{B}$
So, as pressure decreases, volume increases.
b) Want: $\Delta(p) \propto \Delta(d)\implies\Delta(V) \propto \Delta(p)$?
$\Delta(p) \propto \Delta(d)\implies 2\Delta(p) \propto 2\Delta(d)$
$\Delta(V) \propto \Delta(p)\implies 2\Delta(p) \propto 2\Delta(V)$
So, the change in volume scales by 2.
c) Want: $\tfrac{\Delta(V_l)}{\Delta(V_g)}$?
$B_l = (0.25)B_g \Leftrightarrow B_g = 4B_l$
$Recall: \Delta(V) = -kV_o\Delta(p), k=\tfrac{1}{B}$
$\Delta(V_l) = -\tfrac{V_o\Delta(p)}{B_l}, \Delta(V_g) = -\tfrac{V_o\Delta(p)}{B_g}$
$ \implies \tfrac{\Delta(V_l)}{\Delta(V_g)}=(-\tfrac{V_o\Delta(p)}{B_l})(-\tfrac{B_g}{V_o\Delta(p)}), but B_g=4B_l$
$ \implies \tfrac{\Delta(V_l)}{\Delta(V_g)}=(-\tfrac{V_o\Delta(p)}{B_l})(-\tfrac{4B_l}{V_o\Delta(p)})=4$
$\implies\Delta(V_l)=4\Delta(V_g)$
So, the change of volume for lead is four times greater than the change of volume for gold.
Work Step by Step
$\because$ means "because"
$\implies$ means "implies"
$\Leftrightarrow$ means "is equivalent to"
$\propto$ means "is proportional to"
