Explanation:

Let $T_d$ be the doubling time, which is $3$ minutes. Let $C$ be the full capacity of the vessel. \nFor Vessel B:\nInitial number of amoebae ($N_{0,B}$) = $2$.\nTime to fill ($T_B$) = $3$ hours $= 3 \times 60 = 180$ minutes.\nAt time $T_B$, Vessel B is full, so it contains $C$ amoebae. The number of amoebae at time $t$ is given by $N(t) = N_0 \times 2^{t/T_d}$.\nSo, $C = 2 \times 2^{180/3} = 2 \times 2^{60} = 2^{61}$. This is the capacity of the vessel. \nFor Vessel A:\nInitial number of amoebae ($N_{0,A}$) = $1$.\nWe want to find the time ($T_A$) when Vessel A gets filled, i.e., when $N_A(T_A) = C$.\nUsing the formula for Vessel A: $C = 1 \times 2^{T_A/T_d}$\nSubstitute the value of $C$ and $T_d$: $2^{61} = 2^{T_A/3}$\nEquating the exponents: $61 = \frac{T_A}{3}$ $T_A = 61 \times 3 = 183$ minutes. \nTo convert $183$ minutes into hours and minutes: $183 \text{ minutes} = 3 \text{ hours and } 3 \text{ minutes}$ (since $3 \times 60 = 180$). \nAlternatively, consider the relative starting amounts:\nVessel A starts with $1$ amoeba, and Vessel B starts with $2$ amoebae. So, Vessel A starts with half the number of amoebae as Vessel B.\nIf Vessel B is full at $180$ minutes, it means $3$ minutes before that (at $177$ minutes), it was half full (because amoebae double every $3$ minutes).\nSince Vessel A starts with half the initial amount of Vessel B, at any given time, Vessel A will have half the number of amoebae compared to Vessel B.\nTherefore, at $180$ minutes, when Vessel B is full (has $C$ amoebae), Vessel A will have $C/2$ amoebae (i.e., it will be half full).\nSince amoebae double every $3$ minutes, if Vessel A is half full at $180$ minutes, it will become fully filled $3$ minutes later.\nSo, Vessel A will be filled at $180 + 3 = 183$ minutes, which is $3$ hours and $3$ minutes. \nLet's evaluate the options:\nOption (1) $3$ hours ($180$ minutes): Incorrect, as A starts with half the amount of B.\nOption (2) $2$ hours $57$ minutes ($177$ minutes): Incorrect, this is when B was half full.\nOption (3) $3$ hours $3$ minutes ($183$ minutes): This matches our calculation.\nOption (4) $6$ hours ($360$ minutes): Incorrect. \nTherefore, Vessel A will get filled in $3$ hours $3$ minutes. The correct option is (3) 3 hours 3 minutes.