A voltmeter with internal resistance of x\ can be used to measure...

Let the maximum current for full-scale deflection of the voltmeter be $I_g$.

$I_g = \dfrac{V_1}{R_v} = \dfrac{20}{x}$

To increase the measuring range to $V_2 = 30$ V, a resistance $R$ must be connected in series with the voltmeter.

$V_2 = I_g(R_v + R)$

$30 = \dfrac{20}{x}(x + R)$

$30x = 20x + 20R$

$10x = 20R$

$R = \dfrac{x}{2}\ \Omega$

Thus, a resistor of $\dfrac{x}{2}\ \Omega$ should be connected in series with the voltmeter.

Answer: connect resistor of $\dfrac{x}{2}\ \Omega$, in series with voltmeter.