Answer
a) $A=0.0065m^2$
b) $V_{max}=7.8KV$
Work Step by Step
(a) We know that
$A=\frac{Cd}{k\epsilon_{\circ}}$
We plug in the known values to obtain:
$A=\sqrt{\frac{22\times 10^{-12}(2.6\times 10^{-3})}{(1.0059)(8.85\times 10^{-12})}}$
$A=0.0065m^2$
(b) As we know that
$V_{max}=E_{max}d$
We plug in the known values to obtain:
$V_{max}=(3.0\times 10^{6})(2.6\times 10^{-3})$
$V_{max}=7.8KV$