The reading in the ideal voltmeter V shown in the given circuit diagram is:

As the voltmeter is idea, no current will flow through it. Therefore, considering left loop, we can write

$i=\frac{E_{eq}}{r_{eq}}=\frac{8\times 5}{8\times 0.2}$

$\Rightarrow I=25 \mathrm{A}$

Now, potential drop across the cell which is connected in parallel with Voltmeter,

$V=E-ir$

$=5-0.2\times 25$

$=0 \mathrm{V}$