Solution:

The root mean square (r.m.s.) velocity \( v_{\text{rms}} \) of gas molecules is given by:

\[
v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}
\]

For nitrogen (\( \text{N}_2 \)) and oxygen (\( \text{O}_2 \)) gases to have the same \( v_{\text{rms}} \):

\[
\sqrt{\frac{3 k_B T_{\text{N}_2}}{M_{\text{N}_2}}} = \sqrt{\frac{3 k_B T_{\text{O}_2}}{M_{\text{O}_2}}}
\]

Squaring both sides and simplifying:

\[
\frac{T_{\text{N}_2}}{T_{\text{O}_2}} = \frac{M_{\text{N}_2}}{M_{\text{O}_2}}
\]

Given:
- \( T_{\text{O}_2} = 127^\circ \text{C} = 127 + 273 = 400 \, \text{K} \),
- Molar masses \( M_{\text{N}_2} = 28 \) and \( M_{\text{O}_2} = 32 \).

Substitute values:

\[
T_{\text{N}_2} = \frac{28}{32} \times 400 = 350 \, \text{K}
\]

Convert \( 350 \, \text{K} \) to Celsius:

\[
350 - 273 = 77^\circ \text{C}
\]

Answer: The temperature is \( 77^\circ \text{C} \).