Explanation: Field intensity (E) is proportional to the square root of transmit power (P), meaning E $\propto$ $\sqrt{P}$. When power is doubled, the new power (P') becomes 2P. The new field intensity (E') will be proportional to $\sqrt{2P}$, which simplifies to $\sqrt{2}$ $\times$ $\sqrt{P}$. So, E' = $\sqrt{2}$ $\times$ E. Decibels for voltage or field intensity (amplitude quantities) are calculated using the formula: dB = 20 $\log_{10}$ (E'/E) Substituting E' = $\sqrt{2}$ $\times$ E: dB = 20 $\log_{10}$ ($\frac{\sqrt{2} \times E}{E}$) dB = 20 $\log_{10}$ ($\sqrt{2}$) dB = 20 $\log_{10}$ ($2^{0.5}$) Using the logarithm rule $\log(a^b) = b \times \log(a)$: dB = 20 $\times$ 0.5 $\times$ $\log_{10}$(2) dB = 10 $\times$ $\log_{10}$(2) Since $\log_{10}$(2) is approximately 0.301: dB $\approx$ 10 $\times$ 0.301 = 3.01 dB. Therefore, doubling the power results in a 3 dB increase in field intensity. Option B (6 dB) would be the change if the field intensity (or voltage) itself was doubled, not the power. Options C and D are incorrect as they do not align with the logarithmic relationship.