Answer
$R=158k\Omega$
Work Step by Step
To have an error of 5%, voltmeter must read
$R_{eq}=(\frac{1}{15k\Omega}+\frac{1}{R})^{-1}+15k\Omega$
$I=\frac{8.0V}{(\frac{1}{15k\Omega}+\frac{1}{R})^{-1}+15k\Omega}$
$V=\Big(\frac{8.0V}{(\frac{1}{15k\Omega}+\frac{1}{R})^{-1}+15k\Omega}\Big)\Big((\frac{1}{15k\Omega}+\frac{1}{R})^{-1}\Big)$
$\frac{\Big(\frac{8.0V}{(\frac{1}{15k\Omega}+\frac{1}{R})^{-1}+15k\Omega}\Big)\Big((\frac{1}{15k\Omega}+\frac{1}{R})^{-1}\Big)-4.0V}{4.0V}\times100\%=\pm5\%$
$c=(\frac{1}{15k\Omega}+\frac{1}{R})^{-1}$
$\Big(\frac{8.0V}{c+15k\Omega}\Big)c=4.2$
$c=0.525c+7875$
$c=16600$
$c=(\frac{1}{15k\Omega}+\frac{1}{R})^{-1}=16600$
$R=158k\Omega$
