Answer
$(B)~1.22V$
Work Step by Step
We can find the required voltage as follow:
$I=\frac{\mathcal{E}}{R+R_{pf}}$
We plug in the known values to obtain:
$I=\frac{50.0V}{0.1\times 10^7\Omega+4\times 10^7\Omega}$
$I=1.22\times 10^{-6}A$
Now $V=IR$
We plug in the known values to obtain:
$V=1.22\times 10^{-6}A\times 10^6\Omega$
$V=1.22V$
