Answer
a) $9.52\times10^6 \;\rm Pa$
b) See the graph below.
Work Step by Step
a) Since the gas undergoes an isochoric process, so its volume remains constant.
And we know, for an ideal gas, that
$$\dfrac{P_1 \color{red}{\bf\not} V_1}{T_1}=\dfrac{P_2 \color{red}{\bf\not} V_2}{T_2}$$
in our case, the volume is constant.
$$\dfrac{P_1 }{T_1}=\dfrac{P_2 }{T_2}$$
We need to find the final pressure of the gas, so we need to solve for $P_2$
$$P_2=\dfrac{P_1T_2}{T_1 }\tag 1$$
Now we need to find the initial pressure $P_1$ which is given
$$P_1V=nRT_1$$
$$P_1=\dfrac{nRT_1}{V}\tag 2$$
Plugging into (1);
$$P_2=\dfrac{nR \color{red}{\bf\not} T_1}{V}\dfrac{ T_2}{ \color{red}{\bf\not} T_1 }$$
$$P_2=\dfrac{nRT_2}{V} $$
Plugging the known;
$$P_2 =\dfrac{(0.1)(8.31)(300+273)}{(50\times 10^{-6})} $$
$$P_2=\color{red}{\bf 9.52\times10^6}\;\rm Pa$$
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b) To draw a proper scale graph, we need to find the magnitude of the initial pressure.
Plugging the known into (2);
$$P_1=\dfrac{(0.1)(8.31)(20+273)}{(50\times 10^{-6})}=\bf 4.87\times 10^6\;\rm Pa$$
See the figure below.