Answer
$R=100\Omega$.
Work Step by Step
Assume the switches are both open. The circuit is a simple series circuit. Use Ohm’s Law to find the current.
$$I=\frac{6.0V}{50\Omega +20 \Omega + 10\Omega}=0.075 A$$
Now assume the switches are both closed. The $20 \Omega$ resistor is in parallel with the unknown R. Use Kirchhoff’s loop rule on the outer loop of the circuit, using the current through the $20 \Omega$ resistor that we just found.
$$6.0V-I(50\Omega)-(0.075)(20 \Omega)=0$$
$$I=0.090A$$
This is the current in the parallel combination of the $20 \Omega$ resistor and the unknown R. Since 0.075 A flows in the $20 \Omega$ resistor, 0.015 A must flow through R. The voltage drop across R is the same as the drop across the $20 \Omega$ resistor because they are in parallel.
$$V_{20}=(0.075A)(20\Omega)=IR=(0.015A)(R)$$
$$R=100\Omega$$