Answer
$R$ = $ 4 $ $ohms$
Work Step by Step
First, let us compute the equivalent resistance $(R_{eq})$ seen by the source. (NOTE: Please refer to the attached diagram for the computation of $(R_{eq}$)
$R_{5}$ and $R_{6}$ in parallel: $R_{56}$ $=$ $\frac{4\times4}{4+4}$ $=$$ 2 $ $ ohms $
$R_{4}$ and $R_{56}$ in series: $R_{456}$ $=$ $2+6$ $=$$ 8 $ $ ohms $
$R_{456}$ and $R_{7}$ in parallel: $R_{4567}$ $=$ $\frac{8\times24}{8+24}$ $=$$ 6 $ $ ohms $
$R_{3}$ and $R_{4567}$ in series: $R_{34567}$ $=$ $4+6$ $=$$ 10 $ $ ohms $
$R_{2}$ and $R_{34567}$ in parallel: $R_{234567}$ $=$ $\frac{15\times10}{15+10}$ $=$$ 6 $ $ ohms $
$R$ and $R_{234567}$ in series: $R_{eq}$ $=$ $6$ $+$ $R$ $ohms $
Given the power absorbed by $R_{2}$, we can solve for the voltage across it:
$P_{R_{2}}$ $=$ $\frac{(V_{R_{2}})^{2}}{R_{2}}$
$15$ $=$ $\frac{(V_{R_{2}})^{2}}{15}$
$(V_{R_{2}})^{2}$ $=$ $15\times15$
$V_{R_2}$ $=$ $\sqrt 225$
$V_{R_2}$ $=$ $15$ $volts$
and the current $I_{R_{2}}$ passing through $R_{2}$ is:
$I_{R_2}$ $=$ $\frac{V_{R_2}}{R_{2}}$ $=$ $\frac{15}{15}$ $=$ $1$ $ampere$
Using $V_{R_{2}}$, we can solve $V_{R_{3}}$ using voltage divider:
$V_{R_{3}}$ $=$ $V_{R_{2}}$ $\times\frac{R_{3}}{R_{3}+R_{4567}}$ $=$ $15$ $\times\frac{4}{4+6}$ $=$$6$ $volts$
and the current $I_{R_{3}}$ passing through $R_{3}$ is:
$I_{R_{3}}$ $=$ $\frac{V_{R_{3}}}{R_3}$ $=$ $\frac{6}{4}$ $=$ $1.5$ $amperes$
Using $I_{R_{2}}$ and $I_{R_{3}}$, we can now solve for $I_{R_{1}}$ or the total circuit current $I_{T}$:
$I_{T}$ $=$ ${I_{R_{2}} +I_{R_{3}}}$ $=$ $1+1.5$ $=$ $2.5$ $amperes$
Now using the source voltage $V_{T}$ and $V_{R_{2}}$, we can compute for the voltage across $V_{R}$:
$V_{T}$ $=$ $V_{R}$ $+$ $V_{R_{2}}$
$25$ $=$ $V_{R}$ + $15$
$V_{R}$ $=$ $10$ $volts$
Now that we have $V_{R}$ and $I_{R}$, we can now compute for $R$ using ohm's law:
$I_{R}$ $=$ $\frac{V_{R}}{R}$ $=$ $\frac{10}{2.5}$ $=$ $4$ $ohms$
