Can we say DFA is more efficient than NFA?
There are two answers, depending on how you define efficient.

Compactness of representation

Telling more with less: NFAs are more efficient. 

Converting a DFA to an NFA is straightforward and does not increase the size of the representation. 

However, there are regular languages for which the smallest DFA is exponentially bigger than the smallest NFA. A typical example is $(a|b)^*b(a|b)^k$ for $k$ fixed.

Computation

Running it fast: DFAs are more efficient.

The computers we use today are deterministic in nature. That makes them bad at dealing with non-determinism. There are two common ways of dealing deterministically with NFAs: backtracking on one side, which is rather costly, or keeping track of the active states, which means each transition will take up to $N$ times longer (where $N$ is the size of the NFA).