Answer
$5.25\ M$
Work Step by Step
Working from the titration backwards:
Number of moles of $Ag^+$ required:
$0.100\ mol/L\times26.25\ mL\div 1000mL/L=0.002625\ mol$
Which is the same number of moles of $Cl^-$ in the 50.0 mL sample:
$0.002625\ mol\div50.0\ mL\times1000mL/L=0.0525\ M$
Step c is a 10 times dilution, so the concentration from b is:
$10\times0.0525\ M=0.525\ M$
Similarly from a to b:
$10\times0.525\ M=5.25\ M$
