When both switches are open, no current will flow through R.

$i=\frac{1.5}{300+100+50}$

When both switches are close, 50 ohm resistor become short-circuited.

$i=\left(\frac{1.5}{300+\left({\displaystyle \frac{100R}{100+R}}\right)}\right)\times \left(\frac{R}{100+R}\right)$

In both cases, current should be same through Ammeter.

$\frac{1.5}{300+100+50}=\left(\frac{1.5}{300+\left(\frac{100R}{100+R}\right)}\right)\times \left(\frac{R}{100+R}\right)$

$\Rightarrow R=600\Omega$