Answer
a) $1.22\;\rm MPa$
b) See the graph below.
Work Step by Step
a) Since the gas undergoes an isothermal process, its temperature remains constant.
And we know, for an ideal gas, that
$$\dfrac{ P_1V_1}{\color{red}{\bf\not} T_1}=\dfrac{ P_2V_2}{\color{red}{\bf\not} T_2}$$
in our case, the temperature is constant.
$$P_1V_1=P_2 V_2$$
We need to find the final volume of the gas, so we need to solve for $P_2$
$$P_2=\dfrac{ P_1V_1}{V_2}\tag 1$$
Now we need to find the magnitude of the initial pressure which is given by
$$P_1V_1=nRT_1$$
Hence,
$$P_1=\dfrac{nRT_1}{V_1}$$
Plugging into (1);
$$P_2=\dfrac{nRT_1}{ \color{red}{\bf\not} V_1}\dfrac{ \color{red}{\bf\not} V_1}{V_2} $$
$$P_2=\dfrac{nRT_1}{V_2 } $$
Plugging the known;
$$P_2=\dfrac{ (0.1)(8.31)(20+273)}{( 200\times 10^{-6})} $$
$$P_2\approx \color{red}{\bf 1.22\times 10^6}\;\rm Pa$$
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b) To draw a proper scale graph, we need to find the magnitude of the initial pressure (the constant pressure) which is given by
$$P_1V_1=nRT_1$$
Hence,
$$P_1=\dfrac{nRT_1}{V_1}$$
Plugging the known;
$$P_1=\dfrac{(0.1)(8.31)(20+273)}{(50\times 10^{-6})}=\bf 4.87\times 10^6\;\rm Pa$$
See the figure below.
