Answer
$C=0.280pF$
Work Step by Step
We can find the capacitance as
$C=4\pi \epsilon_{\circ}r........eq(1)$
We also know that
$V=\frac{4}{3}\pi r^3$
This simplifies to:
$r=(\frac{3V}{4\pi})^{\frac{1}{3}}$
We plug in the known values to obtain:
$r=(\frac{3(6.7021\times 10^{-8})}{4\times (3.1416)})^{\frac{1}{3}}=0.0025198m$
We plug in the known values in eq(1) to obtain:
$C=4(3.1416)\times (8.85\times 10^{-12})\times (0.0025198)=0.280pF$
