To determine the equivalent noise level, the following formula is used

dB = $20 \times log_{10} \times \frac{P}{P_{o}}$

For 43 dB :-

43 = $20 \times log_{10} \times \frac{P_{1}}{P_{o}}$

43 = $2 \times 10 \times log_{10} \times \frac{P_{1}}{P_{o}}$ = $10 \times log_{10} \times (\frac{P_1}{P_{o}})^2$

4.3 = $log_{10} \times (\frac{P_{1}}{P_{0}})^2$

$10^{4.3} = (\frac{P_{1}}{P_{0}})^2$

For 56 dB :-

56 = $20 log_{10} \times \frac{P_{2}}{P_{o}}$

56 = $2 \times 10 \times log_{10} \times \frac{P_{2}}{P_{o}}$

5.6 = $2 log_{10}(\frac{P_{2}}{P_{0}})$

$10^{5.6} = (\frac{P_{2}}{P_{0}})^2$

For 64 dB :-

64 = $20 log_{10} \times \frac{P_{3}}{P_{o}}$

64 = $2 \times 10 \times log_{10} \times \frac{P_{3}}{P_{o}}$

6.4 = $2 log_{10}(\frac{P_{3}}{P_{0}})$

$10^{6.4} = (\frac{P_{3}}{P_{0}})^2$

For 70 dB :-

70 = $20 log_{10} \times \frac{P_{4}}{P_{o}}$

70 = $2 \times 10 \times log_{10} \times \frac{P_{4}}{P_{o}}$

7 = 2$log_{10}(\frac{P_{4}}{P_{0}})$

$10^{7} = (\frac{P_{4}}{P_{0}})^2$

For 69 dB :-

69 = $20 log_{10} \times \frac{P_{5}}{P_{o}}$

69 = $2 \times 10 \times log_{10} \times \frac{P_{5}}{P_{o}}$

6.9 = 2$log_{10}(\frac{P_{5}}{P_{0}})$

$10^{6.9} = (\frac{P_{5}}{P_{0}})^2$

For 55 dB :-

55 = 20$log_{10} \times \frac{P_{6}}{P_{o}}$

55 = $2 \times 10 \times log_{10} \times \frac{P_{6}}{P_{o}}$

5.5 = 2$log_{10}(\frac{P_{6}}{P_{0}})$

$10^{5.5} = (\frac{P_{6}}{P_{0}})^2$

$\frac{P}{P_{0}} = [(\frac{P_{1}}{P_{0}})^2 + (\frac{P_{2}}{P_{0}})^2 + (\frac{P_{3}}{P_{0}})^2 + (\frac{P_{4}}{P_{0}})^2 + (\frac{P_{5}}{P_{0}})^2 + (\frac{P_{6}}{P_{0}})^2]^{1/2}$

= $[10^{4.3} + 10^{5.6} + 10^{6.4} + 10^{7} + 10^{6.9} + 10^{5.5}]^{1/2}$

= 20 X 3.66

=> Equivalent noise level = 73.2 dB.