Answer
$T^{}= 67.0^{∘}C $
Work Step by Step
The air that strikes the probe will be brought to a complete stop. The thermometer will sense the temperature of this stagnated air, which is the stagnation temperature. The actual air temperature is determined from $$
T=T_0-\frac{V^2}{2 c_p}\\=85^{\circ} \mathrm{C}-\frac{(190 \mathrm{~m} / \mathrm{s})^2}{2 \times 1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}}\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^2 / \mathrm{s}^2}\right)\\=67.0^{\circ} \mathrm{C}
$$
