To solve this problem, we need to find the resonance frequency of a series LCR circuit.$\\$The resonance frequency $f_0$ in a series LCR circuit is given by the formula:$\\ f_0 = \frac{1}{2\pi \sqrt{LC}} \\$Given:$\\$• Inductance $L = 10$ mH$= 10 \times 10^{-3}$ H$\\$• Capacitance $C = 1 \mu$F$= 1 \times 10^{-6}$ F$\\$Substitute the given values into the formula:$\\ f_0 = \frac{1}{2\pi \sqrt{(10 \times 10^{-3})(1 \times 10^{-6})}} \\$Calculate the expression inside the square root:$\\ LC = (10 \times 10^{-3})(1 \times 10^{-6}) = 10 \times 10^{-9}$ H$\cdot$F$\\$Now, calculate the square root:$\\ \sqrt{LC} = \sqrt{10 \times 10^{-9}} = \sqrt{10} \times 10^{-4.5} \\$Calculate $\sqrt{10} \approx 3.162 \\$So, $\sqrt{LC} \approx 3.162 \times 10^{-4.5} \\$Now, substitute back into the formula for $f_0: \\ f_0 = \frac{1}{2\pi \times 3.162 \times 10^{-4.5}} \\$Calculate the denominator:$\\ 2\pi \approx 6.283 \\ 2\pi \times 3.162 \approx 19.84 \\$So, $f_0 \approx \frac{1}{19.84 \times 10^{-4.5}} \\$Calculate $f_0: \\ f_0 \approx \frac{1}{19.84} \times 10^{4.5} \\ f_0 \approx 0.0504 \times 10^{4.5} \\ f_0 \approx 504 \times 10^{1.5} \\ f_0 \approx 504 \times 31.62 \\ f_0 \approx 15900$ Hz$\\$Convert to kHz:$\\ f_0 \approx 15.9$ kHz$\\$Therefore, the frequency at which resonance occurs is $15.9$ kHz.$\\$This corresponds to Option 4.