How to calculate noise voltage after amplification?
Noise figure is defined as:
$$NF = 10log_{10}(\frac{SNR_i}{SNR_o})$$
Thus,
$$5 = 10log_{10}(\frac{V_i^2N_o}{V_o^2N_i})$$
$$5 = -20log_{10}(\frac{V_o}{V_i})+10log_{10}(\frac{N_o}{N_i})$$
First term is the gain, so
$$10log_{10}(\frac{N_o}{N_i}) = 5+60 = 65$$
$$10log_{10}(N_o) = 10log_{10}(N_i) + 65 dB$$
$$10log_{10}(N_o) = 10log_{10}(4kTRB) + 65 dB$$
Hopefully, the symbols are self explanatory.