Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the frequency is given by
$$f=\dfrac{1}{T}$$
where $T$ from the given graph is 0,04 s.
$$f=\dfrac{1}{0.04}=\color{red}{\bf 25}\;\rm Hz$$
$$\color{blue}{\bf [b]}$$
According to Ohm's law,
$$V_R=IR$$
so, the resistance is
$$R=\dfrac{V_R}{I}$$
At $t=0$ s, from the given graph,
$$R=\dfrac{10}{0.5}=\color{red}{\bf 20}\;\rm \Omega$$
$$\color{blue}{\bf [c]}$$
We know that the emf is given by
$$\varepsilon=\varepsilon_0\cos\phi $$
where
$$\phi=\omega t=2\pi f t=2\pi (25)(0.015) =\bf \dfrac{3\pi}{4}\;\rm rad$$
Converting to degrees,
$$\phi = \frac{3\pi}{4}\;\rm rad\times \dfrac{180^\circ}{\pi\;\rm rad}=\bf 135^\circ$$
This is the phase angle for both the voltage and the current at $t=15$ ms which means that the current and the voltage across the resistor are in phase.
So,
$$\varepsilon=\varepsilon_0\cos\phi=10\cos135^\circ=\bf-7.07 \;\rm V$$
And hence,
$$I=I_0\cos135^\circ=0.5\cos135^\circ=\bf -0.354\;\rm A$$
See the figure below.
