Answer
$R_x = 7.5 \mathrm{~\Omega}$
Work Step by Step
The 4 resistors are connected in parallel, therefore, the Ohmmeter measures 2$\Omega$ which is the equivalent resistance of the total resistors. In this case, we could find the resistance $x$ by
\begin{gather*}
\dfrac{1}{R_{\text{eq}}} = \dfrac{1}{R_{1}} + \dfrac{1}{R_{2}} + \dfrac{1}{R_{3}} +\dfrac{1}{R_{x}} \\
\dfrac{1}{R_{x}} = \dfrac{1}{R_{\text{eq}}} - \dfrac{1}{R_{1}} - \dfrac{1}{R_{2}} - \dfrac{1}{R_{3}}
\end{gather*}
Subsitute the values for $R_{1}, R_{2}, R_{3}$ and $R_{\text{eq}}$ to get $R_{x}$
\begin{align*}
\dfrac{1}{R_{x}} & = \dfrac{1}{R_{\text{eq}}} - \dfrac{1}{R_{1}} - \dfrac{1}{R_{2}} - \dfrac{1}{R_{3}} \\
& = \dfrac{1}{ 2 \mathrm{~\Omega} } - \dfrac{1}{ 15 \mathrm{~\Omega}} - \dfrac{1}{ 5 \mathrm{~\Omega}} - \dfrac{1}{ 10 \mathrm{~\Omega}} \\
&= \dfrac{2}{15}
\end{align*}
Therefore, the resistance $R_x$ is the
$$R_x = \dfrac{15}{2} = 7.5 \mathrm{~\Omega}$$
