Identify the element for which electronic configuration in +3 oxidation state is [Ar]3d$$^5$$:

We have to find that element whose electronic configuration, after it has lost three electrons (that is, in the $$+3$$ oxidation state), becomes $$[Ar]\,3d^{5}$$.

First, recall the ground-state electronic configurations of the first-row transition elements. They all possess the $$[Ar]$$ core, followed by a certain number of $$3d$$ electrons and then the $$4s$$ electrons. In general the pattern is

$$$\text{Element (Z)}:\;[Ar]\,3d^{n}\,4s^{2}\quad\text{(with an occasional }4s^{1}\text{ for Cr and Cu)}.$$$

Next, remember the rule for removing electrons to form cations. We always state this rule explicitly:

Electrons are removed first from the outermost shell, so the order of removal for these elements is

$$4s \longrightarrow 3d.$$

Now, let the neutral atom of our unknown element have the configuration

$$[Ar]\,3d^{n}\,4s^{2}.$$

In forming the $$+3$$ ion, we must remove three electrons. Because the two $$4s$$ electrons are higher in energy, they come out first, and one more electron must then be removed from the $$3d$$ subshell. So after ionization the configuration will be

$$[Ar]\,3d^{\,n-1}.$$

We are told that this final configuration must equal $$[Ar]\,3d^{5}.$$ Hence we can equate the exponents:

$$n-1 = 5.$$

Solving for $$n$$ gives

$$n = 6.$$

So the neutral atom must originally contain $$6$$ electrons in the $$3d$$ subshell and $$2$$ electrons in the $$4s$$ subshell. Its ground-state configuration is therefore

$$[Ar]\,3d^{6}\,4s^{2}.$$

Now we consult the periodic table or simply recall that the element with atomic number $$26$$, iron, Fe, has exactly this configuration. Let us verify:

$$\text{Fe (Z = 26)}: [Ar]\,3d^{6}\,4s^{2}.$$

Removing the two $$4s$$ electrons and one $$3d$$ electron gives

$$\text{Fe}^{3+}: [Ar]\,3d^{5},$$

which matches the required configuration.

For completeness we can briefly check the other listed elements:

Ru (Z = 44) lies in the 4d series and would not give a 3d configuration after ionization. Mn (Z = 25) is $$[Ar]\,3d^{5}\,4s^{2}$$; its $$+3$$ state becomes $$[Ar]\,3d^{4}$$, not $$3d^{5}$$. Co (Z = 27) is $$[Ar]\,3d^{7}\,4s^{2}$$; its $$+3$$ state becomes $$[Ar]\,3d^{6}$$. Only Fe (Z = 26) satisfies the condition.

Hence, the correct answer is Option D.