The temperature at which $k_{1}=k_{2}$ will be
$10^{16} e^{-2000 / T}=10^{15} e^{-1000 / T}$
$\Rightarrow \frac{e^{-20001 T}}{e^{-1000 T}}=\frac{10^{15}}{10^{16}}$
$\Rightarrow e^{\frac{-1000}{T}}=10^{-1} \Rightarrow \log _{e} e^{\frac{-1000}{T}}=\log _{e} 10^{-1}$
$\Rightarrow 2.303 \log _{10} e^{\frac{-1000}{T}}=2.303 \times \log _{10} 10^{-1}$
$\Rightarrow \frac{-1000}{T} \times \log _{10} e=-1$ $\Rightarrow T=\frac{1000}{2.303} \mathrm{K}$
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