How to obtain 3db frequency from transfer function
What do you mean by the 3 dB frequency? The frequency at which your magnitude has the value 3 dB? Or are you confusing this with the corner frequency where the magnitude is 3 dB less? 



This would be the frequency marked with the red line in the bode plot.

As -3 dB is corresponding to 

$$20 \cdot log(\frac{1}{\sqrt{2}}) = - 3.01$$

you can infer that

$$20 \cdot log(|H(j \omega)|) = - 3.01$$

$$20 \cdot log(\frac{1}{\sqrt{1+\left(\frac{250}{\omega}\right)^2}}) = - 3.01$$

so 

$$\frac{250}{\omega} = 1$$

hence 

$$\omega = 250$$

Otherwise you could bring the transfer function in a (at least to me) more recognizeable form:

$$H(s) = \frac{s}{s+250}$$

with $$s = j \omega$$

rearranging to

$$H(s) = \frac{\frac{1}{250}s}{s\frac{1}{250}+1}$$

$$H(s) = \frac{\frac{1}{250}s}{s \cdot T+1}$$

where 1\T is the corner frequency (the one you are supposedly be looking for). This kind of notation can vary from textbook to textbook or your teacher. You may also keep writing  $$j \omega$$ and the arrive at $$j \frac{\omega}{\omega_c}$$ 
where $${\omega_c}$$ again is the corner frequency and you can easily read that it's 250.