When the air inside a closed pipe is at a temperature of 20^ ^ ^ C...

The fundamental frequency of a closed pipe is given by $f = \dfrac{v}{4L}$.

The speed of sound in a gas is proportional to the square root of its absolute temperature, $v \propto \sqrt{T}$.

Therefore, the fundamental frequency is proportional to the square root of the absolute temperature, $f \propto \sqrt{T}$.

Given $T_1 = 20^{{}^{{}^{\circ}}}\text{C} = 293\text{ K}$ and $f_1 = 293\text{ Hz}$.

The new temperature is $T_2 = 22^{{}^{{}^{\circ}}}\text{C} = 295\text{ K}$.

Using the relation $\dfrac{f_2}{f_1} = \sqrt{\dfrac{T_2}{T_1}}$:

$f_2 = 293 \sqrt{\dfrac{295}{293}} = \sqrt{293 \times 295}$

$f_2 = \sqrt{(294 - 1)(294 + 1)} = \sqrt{294^2 - 1}$

$f_2 \approx 294\text{ Hz}$