Solution:
NaBH4
Let accept oxidation number of B is x.
\[\begin{array}{*{35}{l}}
Oxidation\text{ }number\text{ }of\text{ }Na\text{ }=\text{ }+1 \\
~ \\
Oxidation\text{ }number\text{ }of\text{ }H\text{ }=\text{ }-\text{ }1 \\
\end{array}\]
Then, at that point, we have:
\[\begin{array}{*{35}{l}}
1\left( +1 \right)\text{ }+\text{ }1\left( x \right)\text{ }+\text{ }4\left( -\text{ }1 \right)\text{ }=\text{ }0 \\
~ \\
=\text{ }1\text{ }+\text{ }x\text{ }-\text{ }4\text{ }=\text{ }0 \\
~ \\
=\text{ }x\text{ }-\text{ }3\text{ }=\text{ }0 \\
~ \\
=\text{ }x\text{ }=\text{ }+3 \\
\end{array}\]
Consequently, Oxidation number of B is +3.