Answer
a) $\approx 813\;\rm kg$
b) $2.54 \;\rm MCi$
Work Step by Step
a) First we need to find the energy produced in one year by converting the power to energy;
$$E=P_Wt\tag 1$$
Now we need to find the mass of $\rm ^{235}_{92}U$ that will operate the power planet for one year since we know that one atom fission gives an energy of 200 MeV.
So we can find the number of $\rm ^{235}_{92}U$ in the year;
$$N_{\rm ^{235}_{92}U}=\dfrac{P_Wt}{200\;\rm MeV}$$
$$N_{\rm ^{235}_{92}U}=\dfrac{2100\times 10^6\times 365.25\times 86400}{200\times 10^6\times 1.6\times 10^{-19}}=\bf 2.071\times 10^{27}\;\rm neuclei$$
$$m=NM=2.071\times 10^{27}\times 235\times 1.67\times 10^{-27}$$
$$m=\color{red}{\bf812.7}\;\rm kg$$
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b) Since $6\%$ of $\rm ^{235}_{92}U$ produces $\rm ^{90}_{38}Sr$,
$$N_{\rm ^{90}_{38}Sr}=0.06N_{\rm ^{235}_{92}U}= 0.06\times 2.071\times 10^{27}\;\rm neuclei$$
$$N_{\rm ^{90}_{38}Sr}= \bf 1.243\times 10^{26}\;\rm neuclei$$
Thus,
$$R=N_{\rm ^{90}_{38}Sr}\lambda=N_{\rm ^{90}_{38}Sr}\dfrac{\ln 2}{T_{_{\frac{1}{2}}}} $$
Plugging the given;
$$R= 1.243\times 10^{26} \dfrac{\ln 2}{29\times 365.25\times 86400} =9.414\times 10^{16}\;\rm decay/s$$
Thus,
$$R=9.414\times 10^{16}\;\rm decay/s\times \dfrac{1\;Ci}{3.7\times 10^{10} \;decay/s}$$
$$=\color{red}{\bf 2.54\times 10^6}\;\rm Ci$$
