Two identical cells each of emf 1 . 5V are connected in series...

Let ${R}_{1} = 10Ω$.

Using $V = iR$

$i = \left(\dfrac{3}{10 + 2r}\right)$

The voltmeter reading is $V ' = 1 . 5V$.

Thus,
$i{R}_{1} = V ' ⇒ 10 \times \dfrac{3}{10 + 2r} = 1 . 5$
$⇒ 20 = 10 + 2r$
$⇒ 2r = 10$
$⇒ r = 5Ω$