Answer
a) $220.4\Omega$
b) $2.009$%
Work Step by Step
a) We are given the potential drop, $V_{f}$, across the filament (let $R$ denote its resistance) as well as the power dissipated by the filament, $P_{f}$. We can hence calculate the resistance $R$ by direct application of the power formula
$$P_{f}=V_{f}I=V_{f}(\frac{V_{f}}{R})=\frac{V_{f}^{2}}{R}$$
$$\Rightarrow{R}=\frac{V_{f}^{2}}{P_{f}}$$
Substituting $V_{f}=115$ and $P_{f}=60$ into the above equation,
$$R=\frac{(115)^{2}}{60}=220.4\Omega$$
b) To calculate the efficiency, we simply compare the output power, $P_{of}$ of the filament to the power dissipated by it. $1W$ of power is equivalent to 680 lumens and it is given that the filament gives off 820 lumens.
$\therefore$ Efficiency $=\frac{P_{of}}{P_{f}}*100=\frac{820(\frac{1}{680})}{60}=2.009$%
