(N/A) No, the bulbs will not glow with the same brightness.
$1$. In a series circuit, the total resistance is $R_{series} = R + R + R = 3R$. Since the voltage $V$ is constant, the current in each bulb is $I_{series} = V / (3R) = (1/3) \times (V/R)$.
$2$. In a parallel circuit, each bulb is connected directly to the source voltage $V$. Therefore, the current in each bulb is $I_{parallel} = V/R$.
$3$. Since the power dissipated by a bulb is $P = I^2R$, the power in the series circuit is $P_{series} = (V/3R)^2 \times R = V^2 / (9R)$, while in the parallel circuit, it is $P_{parallel} = (V/R)^2 \times R = V^2 / R$.
$4$. Because $P_{parallel} > P_{series}$, the bulbs in the parallel circuit will glow much more brightly than those in the series circuit.