Hint: \(N=N_oe^{-\lambda t}\)
Step 1: Find the initial and final number of nuclie in both elements.

$Half-life of{}_{15}{}^{32}P, T_{1/2=}14.3 days$
$Half-life of{}_{15}{}^{33}P, T{\text{'}}_{1/2=}25.3 days and {}_{15}{}^{33}P nucleus decay is 10\% of the total amount of decay.$
$The source has initially 10\% of{}_{15}{}^{33}P nucleus and 90\% of {}_{15}{}^{32}P nucleus.$
$Suppose after t days, the source has 10\% of {}_{15}{}^{32}P nucleus and 90\% of{}_{15}{}^{33}P nucleus.$
$Initially:$
$Number of {}_{15}{}^{33}P nucleus=N$
$Number of {}_{15}{}^{32}P nucleus=9 N$
$Finally:$
$Number of {}_{15}{}^{33}P nucleus = 9 N\text{'}$
$Number of {}_{15}{}^{32}P nucleus =N\text{'}$
${}_{\mathrm{}}{}^{\mathrm{}}$
$$

Step 2: Write the equations for both elements.
For \({ }_{15}^{33} \mathrm{P}\) nucleus, we can write the number ratio as:

$\frac{N\text{'}}{9N}={\left(\frac{1}{2}\right)}^{\frac{1}{T_{1/2}}}$
$N\text{'}=\mathrm{9N}{\left(2\right)}^{\frac{-1}{25.3}}...........................\mathrm{(1)}$
For \({ }_{15}^{32} \mathrm{P}\) nucleus, we can write the number ratio as:
\(\begin{aligned} \frac{9 N^{\prime}}{N} &=\left(\frac{1}{2}\right)^{\frac{1}{T_{1 / 2}}} \\ 9 N^{\prime} &=N(2)^{\frac{-1}{14. 3}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .(2) \end{aligned}\)
$\mathrm{}$
$\mathrm{}\mathrm{}$

Step 3: Find the required time.
On dividing the equation (1) by the equation (2), we get:
\(\frac{1}{9}=9 \times 2\left(\frac{-t}{25.3} +\frac{t}{14.3}\right)\)

$\frac{\mathrm{}}{\mathrm{}}\mathrm{}{\mathrm{}}^{\left(\frac{}{\mathrm{}\mathrm{}}\frac{}{\mathrm{}\mathrm{}}\right)}$
$\frac{1}{81}=2^{\left(\frac{11 t}{25.3\times 14.3}\right)}$
$\log 1- \log 81=\frac{-11t}{25.3\times 14.3}\log 2$
$\frac{-11t}{25.3\times 14.3}=\frac{0-1.908}{0.301}$
$t=\frac{25.3\times 14.3\times 1.908}{11\times 0.301}\approx 208.5 days$
$Hence, it will take about 208.5 days for 90\% decay of{}_{15}{}^{33}P.$